a body thrown vertically upwards with an initial velocity u reaches maximum height in 6second the ratio of distance travelled by the body in for ist and 7 second is
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The ball would reach at top of tower at time,
T = 2u/g = 2*100/10 = 20 s.
that means to cover the height it take 5 seconds.
velocity at the top of tower on returning would be:
v =100 – g*20 = -100 m/s downwards.
Therefore to cover h distance from 2nd eqn. of motion :
h = 100*5 + 0.5*g*25
= 500 + 125
= 625 meter.
T = 2u/g = 2*100/10 = 20 s.
that means to cover the height it take 5 seconds.
velocity at the top of tower on returning would be:
v =100 – g*20 = -100 m/s downwards.
Therefore to cover h distance from 2nd eqn. of motion :
h = 100*5 + 0.5*g*25
= 500 + 125
= 625 meter.
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the ratio is 1:6 , given in the photo
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